Simplifying expression and finding indefinite integral (a) Simplify $$\Large \frac{e^{-4x} + 3e^{-2x}}{e^{-4x}-9} \quad.$$
(b) Hence find $$\Large \int\frac{e^{-4x} + 3e^{-2x}}{e^{-4x}-9} \mathrm{d}x$$
I tried to find a breakdown of the expression, but I have no clue how to simplify that. I can do the integration if I know the simplified expression. 
 A: Hint :
Simplification yields
$$
\frac{e^{-4x} + 3e^{-2x}}{e^{-4x}-9}=\frac{e^{-2x}\left(e^{-2x} + 3\right)}{\left(e^{-2x} + 3\right)\left(e^{-2x} - 3\right)},
$$
where the denominator is using $a^2-b^2=(a+b)(a-b)$. Now, set $u=e^{-2x}-3$ to integrate it.
A: Hint:
$$\frac{e^{-4x}+3e^{-2x}}{e^{-4x}-9} \equiv \frac{e^{-4x}+3e^{-2x}}{e^{-4x}-9} \cdot \underbrace{\left(\frac{e^{4x}}{e^{4x}}\right)}_{1} \equiv \frac{1+3e^{2x}}{1-9e^{4x}} \equiv \require{cancel}\frac{\cancel{1+3e^{2x}}}{\cancel{(1+3e^{2x}})(1-3e^{2x})} \equiv \frac{1}{1-3e^{2x}}.$$
Now, using the substitution $u=1-3e^{2x},$ have a go at integrating this.
A: $$\begin{align}\frac{e^{-4x}+3e^{-2x}}{e^{-4x}-9}=&\frac{e^{-4x}+3e^{-2x}}{e^{-4x}-9}*\frac{e^{+4x}}{e^{+4x}}\\&=\frac{1+3e^{+2x}}{1-9e^{+2x}}\\&=\frac{1+3e^{+2x}}{(1-3e^{+2x})(1+3e^{+2x})}\\&=\frac{1}{(1-3e^{+2x})}\\&=\frac{e^{-2x}}{(e^{-2x}-3)}\\&u=(e^{-2x}-3)\\&du=-2(e^{-2x})dx\\&=\int_{}^{}\frac{-1}{2}\frac{du}{u}\\&=\frac{-1}{2} \ln|u| +c\\&=\frac{-1}{2} \ln|e^{-2x}-3| +c
\end{align}$$
